Get Help with Area Question Homework | Integral of cos2π from 0 to 1/2

[student93]

Homework Statement

Problem is attached in this post.

Homework Equations

Problem is attached in this post.

The Attempt at a Solution

I set my Integral as ∫cos2π dx from 0 to 1/2 and get an answer of 0, which is incorrect.

The correct answer is 1/π, but I don't understand why etc.

 

[Dick]

Homework Statement

Problem is attached in this post.

Homework Equations

Problem is attached in this post.

The Attempt at a Solution

I set my Integral as ∫cos2π dx from 0 to 1/2 and get an answer of 0, which is incorrect.

The correct answer is 1/π, but I don't understand why etc.

If you mean the integral of cos(2πx) from x=0 to x=1/2, then I think the answer is 0. But part of that area is above the x-axis and part is below. I think they want you to add them. Just doing the integral will subtract them.

 

[student93]

If you mean the integral of cos(2πx) from x=0 to x=1/2, then I think the answer is 0. But part of that area is above the x-axis and part is below. I think they want you to add them. Just doing the integral will subtract them.

How exactly would I add them, what method should I use etc.?

 

[SammyS]

How exactly would I add them, what method should I use etc.?

Graph the function from x= 0 to x=1/2, and see what Dick means.

 

[LCKurtz]

@student93: Remember that $\int_a^bf(x)~dx$ is the area under the graph of $y=f(x)$ if $f(x)\ge 0$. The more general form is $$
\text{Area} = \int_a^b y_{upper} - y_{lower}~dx$$The first formula above works because $y_{upper} = f(x)$ and $y_{lower} = 0$ when you are finding the area under a nonnegative function and above the $x$ axis. Do you see why $\int_0^{\frac 1 2} \cos(2\pi x)~dx$ isn't in the form $y_{upper}-y_{lower}$ in the integrand?

 

[student93]

@student93: Remember that $\int_a^bf(x)~dx$ is the area under the graph of $y=f(x)$ if $f(x)\ge 0$. The more general form is $$
\text{Area} = \int_a^b y_{upper} - y_{lower}~dx$$The first formula above works because $y_{upper} = f(x)$ and $y_{lower} = 0$ when you are finding the area under a nonnegative function and above the $x$ axis. Do you see why $\int_0^{\frac 1 2} \cos(2\pi x)~dx$ isn't in the form $y_{upper}-y_{lower}$ in the integrand?

Yes, but then how do I find the area? I'm having trouble setting up an integral that works etc.

 

[Dick]

Yes, but then how do I find the area? I'm having trouble setting up an integral that works etc.

In the interval [0,1/2], where is cos(2*pi*x) positive and where is it negative? Integrate the regions separately and then think how to combine the numbers.

 

[ehild]

The shape is bounded also by the x axis. See picture. Imagine you have to paint the blue area on a wall. You need 1 litre paint to 1 m^2. Calculate how much paint you need. Can you use negative amount of paint?

ehild

 

[LCKurtz]

Yes, but then how do I find the area? I'm having trouble setting up an integral that works etc.

Look at ehild's picture in post #8. Notice that $y_{upper}$ and $y_{lower}$ are two piece functions. Set up two integrals accordingly.